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(C) The angular displacement $	heta$ for a body starting from rest with uniform angular acceleration $\alpha$ is given by $	heta = \frac{1}{2} \alph

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$3 \ 	heta$

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(C) The angular displacement $	heta$ for a body starting from rest with uniform angular acceleration $\alpha$ is given by $	heta = \frac{1}{2} \alpha t^2$.For the first $t_1 = 2 \ s$,the angle rotated is $	heta = \frac{1}{2} \alpha (2)^2 = 2 \alpha$.This implies $\alpha = \frac{	heta}{2}$.For the total time $t_2 = 2 \ s + 2 \ s = 4 \ s$,the total angle rotated is $	heta_{total} = \frac{1}{2} \alpha (4)^2 = 8 \alpha$.The angle rotated in the next $2 \ s$ is $\Delta 	heta = 	heta_{total} - 	heta = 8 \alpha - 2 \alpha = 6 \alpha$.Substituting $\alpha = \frac{	heta}{2}$ into the expression,we get $\Delta 	heta = 6 \left( \frac{	heta}{2} ight) = 3 	heta$.

$A$ disc,initially at rest,is rotated about its axis with a uniform angular acceleration. In the first $2 \ s$,it rotates through an angle $\theta$. In the next $2 \ s$,the disc will rotate through an angle:

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